Graham’s Law of Effusion An unknown gas effuses at a rate that is 0.462 times that of nitrogen gas (at the same temperature). Calculate the molar mass of the unknown gas in g/mol.

Question

Graham’s Law of Effusion

An unknown gas effuses at a rate that is 0.462 times that of nitrogen gas (at the same temperature). Calculate the molar mass of the unknown gas in g/mol.

SORT You are given the ratio of effusion rates
for the unknown gas and nitrogen and asked to find the molar mass of the unknown gas.

GIVEN[latex]\frac{Rate_{unk}}{Rate_{N_{2}}}[/latex]= 0.462
FIND M[latex]_{unk}[/latex]

STRATEGIZE You are given the ratio of rates and you know the molar mass of the nitrogen. You can use Graham’s law of effusion to find the molar mass of the unknown gas.

CONCEPTUAL PLAN

[latex]\frac{Rate_{unk}}{Rate_{N_{2}}}[/latex],M[latex]_{N_{2}}[/latex] → M[latex]_{unk}[/latex]

. [latex]\frac{Rate_{unk}}{Rate_{N_{2}}}=\sqrt{\frac{M_{N_{2}}}{M_{unk}}}[/latex]

RELATIONSHIP USED
[latex]\frac{rate_{A}}{rate_{B}}= \sqrt{\frac{M_{B}}{M_{A}}}[/latex] (Graham’s law)

SOLVE Solve the equation for M[latex]_{unk}[/latex] and substi tute the correct values to calculate it

Accepted Answer

[latex]\frac{Rate_{unk}}{Rate_{N_{2}}}=\sqrt{\frac{M_{N_{2}}}{M_{unk}}}[/latex]

M[latex]_{unk} = \frac{M_{N_{2}}}{(\frac{rate_{unk}}{rate_{N_{2}}})^{2}} =\frac{ 28.02 g/mol}{(0.462)^{2}}[/latex] = 131 g/mol

CHECK The units of the answer are correct. The magnitude of the answer seems reasonable for the molar mass of a gas. In fact, from the answer, you can conclude that the gas is probably xenon, which has a molar mass of 131.29 g/mol.

Question 5.15: Graham’s Law of Effusion An unknown gas effuses at a rate th...

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